# Why is the Fermi level (energy) shfited in doped semiconductors?

the Fermi level is shifted due to doping: Upwards in case of n-type and downwards in case of p-type.

Why is this?

The Fermi level is the level where the probability that an electron occupies the state is $$0.5$$, e.g. where the Fermi energy is located (correct?). I can understand that the distribution changes with the temperatures (it gets broader) but I don't understand why/how the Fermi level changes.

The Fermi energy is defined as:

$$E_F = \frac{h^2}{2m_0} \frac{3 \pi^2 N}{V}^{3/2}$$

I can't see any “doping influence” in this equation. What moves the Fermi energy, and thus the Fermi level?

There are related questions like Why does Fermi Level change due to change in donor atom concentration? and Does the Fermi level depend on temperature? but they do not answer this question here.

• Doping changes $N$ by adding/removing electrons per unit volume when compared to the undoped material (Note: the formula for the Fermi energy you give is not the definition of the Fermi level and only holds for an isotropic, non-interacting system). Aug 17, 2019 at 16:15
• Thank you! So because the number of particles changes it changes the fermi level? That's hard to understand.. What is the definition of the fermi level?
– Ben
Aug 17, 2019 at 16:57
• Ah, I think I got it: The Fermi energy is not related to the Fermi level, right? It is the energy the particle with the highest energy has when the system is in the ground state (so at T=0K always?!). Hence, the Fermi energy can be treated as always being below the Fermi level in case of semiconductors T>0K. Or?
– Ben
Aug 17, 2019 at 17:01